What is the integral from-1 to 1 of 2xe^{-inpix} ?

The integral from-1 to 1 of 2xe^{-inpix} is (4i(-1)^n)/(pin)

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integral from-1 to 1 of 2xe^{-inpix}

\int_{- 1}^{1} 2 x e^{- inπ x} d x

\frac{4 i ( - 1 ) ^{n}}{πn}

Solution steps

\int_{- 1}^{1} 2 x e^{- inπ x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{- 1}^{1} x e^{- inπ x} d x

Simplify

x e^{- inπ x} : x cos (πn x) - i x sin (πn x)

= 2 \cdot \int_{- 1}^{1} x cos (πn x) - i x sin (πn x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int_{- 1}^{1} x cos (πn x) d x - \int_{- 1}^{1} i x sin (πn x) d x)

\int_{- 1}^{1} x cos (πn x) d x = 0

\int_{- 1}^{1} i x sin (πn x) d x = - \frac{2 i ( - 1 ) ^{n}}{πn}

= 2 (0 - (- \frac{2 i ( - 1 ) ^{n}}{πn}))

Simplify

2 (0 - (- \frac{2 i ( - 1 ) ^{n}}{πn})) : \frac{4 i ( - 1 ) ^{n}}{πn}

= \frac{4 i ( - 1 ) ^{n}}{πn}

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What is the integral from-1 to 1 of 2xe^{-inpix} ?
The integral from-1 to 1 of 2xe^{-inpix} is (4i(-1)^n)/(pin)